In the Stability of Minkowski Spacetime, Christodoulou and Klainerman prove a local existence proof for a particular class of quasilinear wave equation for a symmetric, traceless, covariant 2-tensor $k_{ij}$ on $\mathbb{R}^3$. On p.309 (in the print version), the authors claim that a certain term behaves better (in terms of decay at infinity) in the wave equation for $\text{div}(k)$, where $\text{div}$ is the divergence operator on $\mathbb{R}^3$: $\text{div}(k)_j = \nabla^i k_{ij}$. My question is: why? Details below.
In this question I include only the relevant parts of the equation; the irrelevant (for this question) parts will be denoted by $f$. Note that this is a linear wave equation (since it is a part of the iteration argument in the proof of local existence to the quasilinear problem). So suppose we have $$ -(\phi^{-1}\partial_t)^2 k_{ij} + \Delta k_{ij} = \phi^{-2}\nabla_i \nabla_j \dot \phi + f. $$ We are solving for the unknown $k_{ij} : [0, T] \times \mathbb{R}^3 \to \mathbb{R}$, and $\phi : [0, T] \times \mathbb{R}^3 \to \mathbb{R}$ solves $\Delta \phi = |k[n]|^2\phi$. Here, $k[n]$ can be thought of as a fixed 2-tensor of the same type as $k$. (It is denoted as such since this question, again, is concerned with producing a "$k_{n + 1}$" solution to the linear equation above, in a typical hyperbolic equation iteration scheme.) Also, $\dot\phi$ denotes $\partial_t \phi$.
The iteration scheme provides us with the following bounds: $\phi - 1 \in H_{4, -1}, \dot\phi \in H_{4, -1}, k[n] \in H_{3, 1}$. Here $H_{s, \delta}$ is the weighted Sobolev space on $\mathbb{R}^3$: $$ \|{U}\|_{H_{s, \delta}}^2 = \sum_{0 \leq k \leq s} \int_{\mathbb{R}^3} |(1 + |x|^2)^{(\delta + k)/2}\nabla^k U|^2\, dx. $$
The authors claim that $\text{div}(k) \in H_{0, 3}$ and proceed by deriving, from the wave equation for $k$, a wave equation for $\text{div}(k)$ and applying energy estimates using the multiplier vector field $\langle x \rangle^3\partial_t$. In doing so, one must bound (as in standard energy estimates for linear wave equations) $$ \|\langle x \rangle^3\Box \text{div}(k)\|_{L^2_x}. $$ However, if we consider what happens in applying $\text{div}$ to the equation above, we see that we obtain $$ \Box \text{div}(k)_j = 2\phi^{-3}\nabla^i \phi \nabla_i \nabla_j \dot \phi - \phi^{-2}\nabla^i \nabla_i \nabla_j \dot \phi + f. $$ Yet it is straightforward to see that with the estimates given above for $\phi - 1, \dot \phi$, that (for instance) we do not have $\langle x \rangle^3\Box_{g_n} \overline{\text{div}}{(k)_j}\in L^2$. Indeed, the best we can do is place $\phi \in L^\infty$, and then (by weighted Sobolev embeddings) place $\langle x\rangle^{3/2}\nabla \phi \in L^\infty$. This leaves $\|\langle x \rangle^{3/2}\nabla_i \nabla_j \dot \phi\|_{L^2_x}$, but since $\dot \phi \in H_{4, -1} \implies \nabla_i \nabla_j \dot \phi \in H_{2, 1}$ only, this quantity is not bounded.
How is this bound obtained?
